Question

tan^4 + tan^2 theta = sec^4theta - sec^2 theta​

Answer 1

(Instead of θ, I used A in solving this problem)

Step-by-step explanation:

Given :

To prove $Tan^4A + Tan^2A = Sec^4A - Sec^2A$

I am using the identities ,

Sec²A - Tan²A = 1   ...(I)

⇒ Tan²A = Sec²A - 1   (from I)

Proof :

LHS =  $Tan^4A + Tan^2A$

$=(Tan^2A)(Tan^2A + 1)$     (Taking Tan²A as common)

Subsitituting Tan²A = Sec²A - 1,

We get,

$= (Sec^2A - 1 )((Sec^2A - 1) + 1)$

$= (Sec^2A - 1 )(Sec^2A)$

$= Sec^4A - sec^2A$

= RHS